19.7 Electrolytic Cells
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- เผยแพร่เมื่อ 14 ส.ค. 2017
- Struggling with Electrolysis? Chad breaks down the principles of Electrolysis and how to determine the products of both Molten and Aqueous Electrolysis.
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Hi Chad, so I understand reduction is always at the cathode and electrons flow from anode to cathode but how do you know which metal will be undergoing reduction/oxidation? usually, examples have Cu at the cathode and Zn at the anode for voltaic cells, but why couldn't it be the other way around and have Zn undergo reduction? For the DAT they don't give you a table with the Ecell values, sometimes they give reduction potentials but not always. Thanks!
Hi sir,awesome job! Well, i want to make an electrolytic cell with two electrodes of iron (Fe) and the electrolyte of H2SO4. While charging the anode metal goes to FeO2 and cathode to Fe. It's on principle of lead-aced battery. Do you think it can work? I have to try an other type of electrolyte?
Finally understood electrolysis, thank you! Quick question: would the cathode and anode here be inert, like Pt or C graphite?
Exactly right Kool Kola! You definitely want to use inert electrodes, but this can be problematic with electrolysis. For aqueous electrolysis you can use a lot of metals as inert electrodes as they are already in their most reduced state as a metal, but you have to be selective in choosing your anode and platinum is a good choice.
For molten electrolysis you also have to choose metals with higher melting points than the melting point of the salt being electrolysed, otherwise your electrodes will melt too (know someone who ran into this problem!).
Hope this helps!
@@ChadsPrep Thank you so much, this makes sense!
@@SaketRamCoolPolar You are welcome - glad to hear it!
Why is the e cell -3.78? Don’t we do cathode - anode?
If you are provided with REDUCTION potentials then yes. And the reason we subtract the anode is that oxidation takes place there rather than reduction and subtracting changes the sign since it is the reverse reaction that is actually occurring. But if you look back at the video the -1.07V value I provided was actually already for the oxidation (the corresponding reduction potential is +1.07V). So had I provided the reduction potential for you then you would have indeed done "cathode - anode" and you would have calculated -2.71V - 1.07V = -3.78V.
But I already changed the sign for you by providing you with the oxidation potential in which case you simply add the values -2.71V + -1.07V = -3.78V
Same answer either way. I cover the exact reasoning for calculating it either of these two ways a couple lessons earlier here: th-cam.com/video/iBiumUZfKjA/w-d-xo.html
Happy Studying!
@@ChadsPrep Appreciate that!!:-) i think for the dat they only give reduction potentials hence the confusion(at least for the Canadian one!)
How can you tell which one is more positive? I'm still confused.
You have to use the table of standard reduction potentials. You can find it on google if you just type it in.
Which textbook do you use
@@kasitadaniel8716 it might be brown lemay but I might be wrong.